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Acta Mathematica Vietnamica

, Volume 39, Issue 2, pp 121–131 | Cite as

Global Existence and Decay Rates of Solutions of Generalized Benjamin–Bona–Mahony Equations in Multiple Dimensions

  • Hongmei XuEmail author
  • Guyu Xin
Open Access
Article

Abstract

We study the global existence and decay rates of the Cauchy problem for the generalized Benjamin–Bona–Mahony equations in multi-dimensional spaces. By using Fourier analysis, frequency decomposition, pseudo-differential operators and the energy method, we obtain global existence and optimal L 2 convergence rates of the solution.

Keywords

Cauchy problem Generalized Benjamin–Bona–Mahony equation Multiple dimensions Global existence Optimal L2 decay estimate 

Mathematics Subject Classification (2010)

35L05 

1 Introduction

In this paper, we study the global existence and decay rates of the smooth solution u(x,t) to the scalar multi-dimensional generalized Benjamin–Bona–Mahony (GBBM) equations of the form
$$ \begin{cases} \partial_t u-\Delta\partial_t u-\eta\Delta u+(\beta\cdot\nabla)u+\operatorname {div}f(u)=0,\\ u(0,x)=u_0(x). \end{cases} $$
(1.1)
Here η is a positive constant, β is a real constant vector, f(u)=(u 2,u 2,…,u 2), \(\Delta=\sum_{j=1}^{n}\partial_{x_{j}}^{2}\) is the Laplacian, ∇ is a gradient operator, n≥2 is the spatial dimension.
The well known Benjamin–Bona–Mahony (BBM) equation
$$ u_t-u_{xxt}+u_x+uu_x=0, \quad - \infty<x<\infty,\ t>0 $$
and its counterpart, the Korteweg–de Vries (KdV) equation
$$u_t-u_{xxx}+u_x+uu_x=0, \qquad - \infty<x<\infty,\ t>0, $$
were both suggested as model equations for long waves in nonlinear dispersive media. The BBM equation was advocated by Benjamin, Bona and Mahony [3] in 1972. Since then, the periodic boundary value problems, the initial value problems and the initial boundary value problems, for various generalized BBM equations have been studied. The existence and uniqueness of solutions for GBBM have been proved by many authors [2, 3, 7, 8]. The decays of solutions were also studied in [1, 4, 5, 6, 10]. In [1, 4, 5, 10], the authors studied the equation in low spatial dimension. In [6], the equation in high spatial dimension was studied, but the authors just got a global solution u(x,t)∈H 1. In this paper, the goal is to get a global smoother solution and give its L 2 decay rates in high spatial dimension.
Throughout this paper, we denote the generic constants by C and write \(D^{\alpha}f=\partial_{x_{1}}^{\alpha_{1}}\partial_{x_{2}}^{\alpha_{2}}\cdots \partial_{x_{n}}^{\alpha_{n}}\) for a multi-index α=(α 1,α 2,…,α n ). Let \(W^{s,p}(\mathbb{R}^{n})\), sZ +, p∈[1,∞], be the usual Sobolev space with the norm
$$ \|f\|_{W^{s,p}}:=\sum_{|\alpha|=0}^{s} \bigl\| D^{\alpha}f\bigr\| _{L_p}. $$
In particular, W s,2=H s . The Fourier transformation with respect to the variable \(x\in{\mathbb{R}}^{n}\) is
$$ \hat{f}(\xi,t)=\int_{{\mathbb{R}}^n}f(x,t)e^{-\sqrt{-1}x\cdot\xi}\,dx $$
and the inverse Fourier transformation with respect to the variable ξ is
$$ f(x,t)=(2\pi)^{-n}\int_{{\mathbb{R}}^n}\hat{f}( \xi,t)e^{\sqrt {-1}x\cdot\xi}\,d\xi. $$
In this paper, all convolutions are only with respect to the spatial variable x.

Our main result is the following:

Theorem 1.1

If \(E=(\|u_{0}\|_{H^{l}}+\|\bigtriangledown u_{0}\|_{H^{l}})\) is small enough, \(l\geq1+[\frac{n}{2}]\), then there exists a global solution u(x,t) of (1.1) such that
$$ u\in L_{\infty}\bigl(0,\infty;H^l\bigl({\mathbb{R}}^n\bigr)\bigr). $$
Moreover, we have \(\|D_{x}^{\alpha}u\|_{L_{2}}\leq C(1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}\) for |α|≤l.

Remark

The decay rate is the same as that of the heat equation, so our estimate is optimal.

The rest of the paper is arranged as follows. In Sect. 2, we get the local existence of the solution directly by constructing a Cauchy sequence and using energy estimation. In Sect. 3, by means of the Green function, we obtain a bound of the solution, then we extend the local solution to the global one. In Sect. 4, we get an L 2 decay estimate of the solution.

2 Local existence

In this section, we will construct a convergent sequence {u (m)(x,t)} to get the local solution, where u (m)(x,t) satisfy the following linear problem
$$ \begin{cases} \partial_t u^{(m)}-\Delta\partial_t u^{(m)}-\eta\Delta u^{(m)}+(\beta \cdot\nabla)u^{(m)}=-\operatorname {div}f(u^{(m-1)}),\\ u^{(m)}(0,x)=u_0(x),\quad u^{(0)}(t,x)=0. \end{cases} $$
(2.1)
We will construct a Banach space and prove that the sequence is convergent in this space, so the limit is the solution of (1.1).
First, we define a function space
$$ \textbf{X}_{T_0}= \Bigl\{ u(x,t)|\|u\|_{\textbf{X}_{T_0}}=\sup _{0\leq t\leq T_0}\|u\|_{H^l}\leq C_0E \Bigr\} . $$
Here \(E=(\|u_{0}\|_{H^{l}}+\|\nabla u_{0}\|_{H^{l}})\) is small enough, \(C_{0}>\sqrt{l},l\geq1+[\frac{n}{2}]\) and n≥2. The metric in \(\textbf{X}_{T_{0}}\) is induced by the norm \(\|u\|_{\textbf{X}_{T_{0}}}\):
$$ \rho(u,v)=\|u-v\|_{\textbf{X}_{T_0}}\quad\forall u,v\in\textbf{X}_{T_0}. $$
It is obvious that \(\textbf{X}_{T_{0}}\) is a nonempty complete space.

Lemma 2.1

There exists some constant T 1 such that {u (m)(x,t)} belongs to \(\textbf{X}_{T_{1}}\).

Proof

We will prove this lemma by induction on m. When m=1, we have
$$ \begin{cases} u_t^{(1)}-\Delta u_t^{(1)}-\eta\Delta u^{(1)}+(\beta\cdot\nabla )u^{(1)}=0,\\ u^{(1)}(0,x)=u_0(x). \end{cases} $$
(2.2)
Multiplying the first equation by u (1) and integrating with respect to the variable x in \(\mathbb{R}^{n}\), we get
$$ \frac{1}{2}\frac{\partial}{\partial t}\bigl\| u^{(1)}\bigr\| _{L_2}^2+ \frac{1}{2}\frac{\partial}{\partial t}\bigl\| \nabla u^{(1)}\bigr\| _{L_2}^2+ \eta\bigl\| \nabla u^{(1)}\bigr\| _{L_2}^2=0, $$
here we use the fact that ∫(β⋅∇u (1))⋅u (1) dx=0. Then
$$ \bigl\| u^{(1)}\bigr\| _{L_2}^2+\bigl\| \nabla u^{(1)} \bigr\| _{L_2}^2\leq \|u_0\|_{L_2}^2+ \|\nabla u_0\|_{L_2}^2. $$
From (2.2) we have
$$ D_x^{\alpha}u_t^{(1)}-D_x^{\alpha} \Delta u_t^{(1)}-\eta D_x^{\alpha} \Delta u^{(1)}+D_x^{\alpha}(\beta\cdot\nabla)u^{(1)}=-D_x^{\alpha} \operatorname {div}f\bigl(u^{(0)}\bigr)=0. $$
(2.3)
Multiplying the two sides in (2.3) by \(D_{x}^{\alpha}u^{(1)}\) and integrating with respect to the variable x in \(\mathbb{R}^{n}\), we also get
$$ \frac{1}{2}\frac{\partial}{\partial t}\bigl\| D_x^{\alpha}u^{(1)} \bigr\| _{L_2}^2+\frac{1}{2}\frac{\partial}{\partial t} \bigl\| D_x^{\alpha}\nabla u^{(1)}\bigr\| _{L_2}^2 +\eta\bigl\| \nabla D_x^{\alpha}u^{(1)} \bigr\| _{L_2}^2=0 . $$
(2.4)
From (2.4) we have
$$ \bigl\| D_x^{\alpha}u^{(1)}\bigr\| _{L_2}^2(t)+ \bigl\| D_x^{\alpha}\nabla u^{(1)}\bigr\| _{L_2}^2(t) \leq \bigl\| D_x^{\alpha}u_0\bigr\| _{L_2}^2+ \bigl\| D_x^{\alpha}\nabla u_0\bigr\| _{L_2}^2. $$
Thus, for every T 1, we have \(u^{(1)}\in\textbf{X}_{T_{1}}\). Supposing that \(u^{(m)}(x,t)\in\textbf{X}_{T_{1}}\), we next prove that \(u^{(m+1)}(x,t)\in\textbf{X}_{T_{1}}\). By definition, u (m+1)(x,t) satisfies the following equation:
$$ \begin{cases} \partial_t u^{(m+1)}-\Delta\partial_t u^{(m+1)}-\eta\Delta u^{(m+1)}+(\beta\cdot\nabla)u^{(m+1)}=-\operatorname {div}f(u^{(m)}),\\ u^{(m+1)}(0,x)=u_0(x). \end{cases} $$
Then
$$ D_x^{\alpha}\partial_t u^{(m+1)}-D_x^{\alpha} \Delta\partial_t u^{(m+1)}-\eta D_x^{\alpha} \Delta u^{(m+1)}+D_x^{\alpha}(\beta\cdot \nabla)u^{(m+1)}=-D_x^{\alpha} \operatorname {div}f \bigl(u^{(m)}\bigr), $$
so, we get
$$\begin{aligned} \begin{aligned} &\frac{1}{2}\frac{\partial}{\partial t} \bigl\| D_x^{\alpha}u^{(m+1)}\bigr\| _{L_2}^2+ \frac{1}{2}\frac{\partial}{\partial t}\bigl\| D_x^{\alpha}\nabla u^{(m+1)}\bigr\| _{L_2}^2 +\eta\bigl\| \nabla D_x^{\alpha}u^{(m+1)}\bigr\| _{L_2}^2 \\ &\quad =-\int D_x^{\alpha}\text{div}f\bigl(u^{(m)} \bigr)D_x^{\alpha}u^{(m+1)}\,dx \\ &\quad =\int D_x^{\alpha}f\bigl(u^{(m)}\bigr)\nabla D_x^{\alpha}u^{(m+1)}\,dx \\ &\quad \leq\frac{\eta}{2}\bigl\| \nabla D_x^{\alpha}u^{(m+1)} \bigr\| _{L_2}^2+C\bigl\| D_x^{\alpha}f \bigl(u^{(m)}\bigr)\bigr\| _{L_2}^2. \end{aligned} \end{aligned}$$
(2.5)
To proceed, we need the following inequality:
$$ \bigl\| D^s(F G)\bigr\| _{L^r}\leq C\bigl( \bigl\| D^sF\bigr\| _{L^p}\|G\|_{L^q}+\|F\|_{L^p} \bigl\| D^sG\bigr\| _{L^q}\bigr), $$
(2.6)
where 1≤r,p,q≤∞ and \(\frac{1}{r}=\frac{1}{p}+\frac{1}{q}\).
Noticing that f(u)=(u 2,u 2,…,u 2), we have
$$ \bigl\| D_x^{\alpha}f\bigl(u^{(m)}\bigr) \bigr\| _{L_2}\leq C\bigl\| u^{(m)}\bigr\| _{L_{\infty}}\bigl\| D_x^{\alpha}u^{(m)} \bigr\| _{L_2}. $$
(2.7)
By the Sobolev embedding inequality, we have \(\|u\|_{L_{\infty}}\leq C\|u\|_{H^{l}}\) if \(l\geq1+[\frac{n}{2}]\). From (2.5), (2.7), we get
$$\begin{aligned} &\frac{1}{2}\frac{\partial}{\partial t}\bigl\| D_x^{\alpha}u^{(m+1)} \bigr\| _{L^2}^2+\frac{1}{2}\frac{\partial}{\partial t} \bigl\| D_x^{\alpha}\nabla u^{(m+1)}\bigr\| _{L^2}^2 +\frac{\eta}{2}\bigl\| \nabla D_x^{\alpha}u^{(m+1)} \bigr\| _{L^2}^2 \\ &\quad \leq C\bigl\| u^{(m)}\bigr\| _{H^l}^2\bigl\| D_x^{\alpha}u^{(m)} \bigr\| _{L^2}^2. \end{aligned}$$
Then
$$\begin{aligned} \bigl\| D_x^{\alpha}u^{(m+1)} \bigr\| _{L^2}^2+\bigl\| D_x^{\alpha}\nabla u^{(m+1)}\bigr\| _{L^2}^2 \leq&\bigl\| D_x^{\alpha}u_0 \bigr\| _{L^2}^2+\bigl\| D_x^{\alpha}\nabla u_0\bigr\| _{L^2}^2+C\int_0^tE^2E^2\,ds \\ \leq& E^2+CE^4t. \end{aligned}$$
(2.8)

When \(T_{1}\leq(C_{0}^{2}-1)\frac{1}{CE^{2}}\), from (2.8) we have \(\{u^{(m+1)}(x,t)\}\in\textbf{X}_{T_{1}}\). By the induction method, our lemma is proved. □

Lemma 2.2

There exists a positive constant T 0 such that {u (m)(x,t)} constructed by (2.1) is a Cauchy sequence in \(\textbf{X}_{T_{0}}\).

Proof

We only need to prove that
$$ \bigl\| u^{(m+1)}-u^{(m)}\bigr\| _{\textbf{X}_{T_0}}\leq\lambda\bigl\| u^{(m)}-u^{(m-1)}\bigr\| _{\textbf{X}_{T_0}} $$
(2.9)
for some 0<λ<1. By (2.1), for every m, u (m+1)u (m) satisfies the following equation:
$$ \begin{cases} \partial_t (u^{(m+1)}-u^{(m)})-\Delta\partial _t(u^{(m+1)}-u^{(m)})-\eta\Delta(u^{(m+1)}-u^{(m)})\\ \quad{} +(\beta\cdot\nabla)(u^{(m+1)}-u^{(m)}) =-\operatorname {div}f(u^{(m)})+\operatorname {div}f(u^{(m-1)}),\\ (u^{(m+1)}-u^{(m)})(0,x)=0. \end{cases} $$
(2.10)
Thus we have
$$\begin{aligned} &D_x^{\alpha}\partial_t \bigl(u^{(m+1)}-u^{(m)}\bigr)-D_x^{\alpha} \Delta\partial _t\bigl(u^{(m+1)}-u^{(m)}\bigr)-\eta D_x^{\alpha}\Delta\bigl(u^{(m+1)}-u^{(m)} \bigr) \\ &\quad {} +D_x^{\alpha}(\beta\cdot\nabla) \bigl(u^{(m+1)}-u^{(m)} \bigr)=-D_x^{\alpha} \operatorname {div}\bigl(f\bigl(u^{(m)}\bigr)-f \bigl(u^{(m-1)}\bigr)\bigr). \end{aligned}$$
(2.11)
Multiplying the two sides of (2.11) by \(D_{x}^{\alpha}(u^{(m+1)}-u^{(m)})\) and integrating with respect to the variable x in \(\mathbb{R}^{n}\), we get
$$\begin{aligned} &\frac{1}{2}\frac{\partial}{\partial t}\bigl\| D_x^{\alpha} \bigl(u^{(m+1)}-u^{(m)}\bigr)\bigr\| _{L_2}^2+ \frac{1}{2}\frac{\partial}{\partial t}\bigl\| D_x^{\alpha}\nabla \bigl(u^{(m+1)}-u^{(m)}\bigr)\bigr\| _{L_2}^2 +\eta\bigl\| \nabla D_x^{\alpha}\bigl(u^{(m+1)}-u^{(m)} \bigr)\bigr\| _{L_2}^2 \\ &\quad =-\int D_x^{\alpha}{\rm div}\bigl(f\bigl(u^{(m)} \bigr)-f\bigl(u^{(m-1)}\bigr)\bigr) D_x^{\alpha} \bigl(u^{(m+1)}-u^{(m)}\bigr)\,dx\\ &\quad =\int D_x^{\alpha}\bigl(f\bigl(u^{(m)}\bigr)-f \bigl(u^{(m-1)}\bigr)\bigr)\nabla D_x^{\alpha } \bigl(u^{(m+1)}-u^{(m)}\bigr)\,dx \\ &\quad \leq\frac{\eta}{2}\bigl\| \nabla D_x^{\alpha} \bigl(u^{(m+1)}-u^{(m)}\bigr)\bigr\| _{L_2}^2 +C\bigl(\bigl\| u^{(m)}\big\| _{H^l}^2+\bigl\| u^{(m-1)} \bigr\| _{H^l}^2\bigr)\bigl\| D_x^{\alpha} \bigl(u^{(m)}-u^{(m-1)}\bigr)\bigr\| _{L_2}^2. \end{aligned}$$
Thus
$$ \frac{\partial}{\partial t}\bigl(\bigl\| D_x^{\alpha}\bigl(u^{(m+1)}-u^{(m)} \bigr)\bigr\| _{L_2}^2 \leq C\bigl(\bigl\| u^{(m)} \bigr\| _{H^l}^2+\bigl\| u^{(m-1)}\bigr\| _{H^l}^2 \bigr) \bigl(\bigl\| D_x^{\alpha }\bigl(u^{(m)}-u^{(m-1)} \bigr)\bigr\| _{L_2}^2\bigr)\bigr). $$
Noticing that (u (m+1)u (m))(0,x)=0, and choosing \(T_{0}=\min(T_{1}, \frac{1}{4CE^{2}})\), we have
$$\begin{aligned} \bigl\| \bigl(u^{(m+1)}-u^{(m)}\bigr)\bigr\| _{H^l}^2 \leq& C\int_0^t\bigl(\bigl\| u^{(m)} \bigr\| _{H^l}^2+\bigl\| u^{(m-1)}\bigr\| _{H^l}^2 \bigr)\bigl\| u^{(m)}-u^{(m-1)}\bigr\| _{H^l}^2\,ds \\ \leq&2CE^2 T_0 \bigl\| u^{(m)}-u^{(m-1)} \bigr\| _{\textbf{X}_{T_0} }^2 \\ \leq&\frac{1}{2}\bigl\| u^{(m)}-u^{(m-1)}\bigr\| _{\textbf{X}_{T_0} }^2. \end{aligned}$$

Thus we can get (2.9). □

Since \(\textbf{X}_{T_{0}}\) is a complete metric space, by Lemmas 2.1 and 2.2, there exists a \(u(x,t)\in\textbf {X}_{T_{0}}\) such that
$$ \begin{cases} \partial_t u-\Delta\partial_t u-\eta\Delta u+(\beta\cdot\nabla)u+\operatorname {div}f(u)=0,\\ u(0,x)=u_0(x). \end{cases} $$

Thus we have proved the local existence of (1.1). Next we will prove the global existence.

3 Bounded estimates and global existence

In this section, we want to get \(\|u\|\ _{H^{l}}\leq CE\), then extend the local solution to a global one.

Lemma 3.1

If u 0H 1, we have uL (0,+∞;H 1).

Proof

Multiplying the first equation by u and integrating with respect to the variable x in \(\mathbb{R}^{n}\), we get
$$ \partial_tu-\Delta\partial_tu-\eta\Delta u+(\beta\cdot \nabla)u=-\operatorname {div}f(u), $$
then we have
$$ \frac{1}{2}\frac{\partial}{\partial t}\|u\|_{L_2}^2+ \frac{1}{2}\frac{\partial}{\partial t}\|\nabla u\|_{L_2}^2+\eta\| \nabla u\|_{L_2}^2 =-\int u\; \operatorname {div}f(u)\,dx=\int f(u)\cdot \nabla u\,dx=0. $$
Thus
$$ \|u\|_{L_2}^2+\|\nabla u\|_{L_2}^2+ \int_0^t\|\nabla u\|_{L_2}^2 ds\leq \|u_0\|_{L_2}^2+\|\nabla u_0\|_{L_2}^2, $$
(3.1)
and our lemma is proved. □
In order to get a bound of u, first we want to get an explicit expression of u through the Green function. The linearized system of (1.1) is of the form
$$ \partial_t u-\Delta\partial_t u-\eta\Delta u+(\beta\cdot\nabla)u=0. $$
(3.2)
The Green function of (3.2) is defined as
$$ \begin{cases} \partial_t G-\Delta\partial_t G-\eta\Delta G+(\beta\cdot\nabla)G=0,\\ G|_{t=0}=\delta(x). \end{cases} $$
Direct calculation shows that the Fourier transformation of G is
$$ \hat{G}=e^{-\frac{\eta|\xi|^2}{1+|\xi|^2}t-\frac{i\beta\cdot\xi t}{1+|\xi|^2}}. $$
(3.3)
Set
$$ \hat{H}=\frac{1}{1+|\xi|^2}\hat{G}. $$
(3.4)
Due to Duhamel’s principle, we know that the solution of (1.1) can be expressed by
$$ u(x,t)=G\ast u_0-\int_0^t H(t-s)\ast \operatorname {div}f(u) (s)\,ds. $$
(3.5)
Next we want to analyze u through estimates of G,H. For this part, we must estimate decay rates of \(\hat{G}, \hat{H}\) separately for low and high frequencies. Thus we set
$$ \chi_1(\xi)= \begin{cases} 1, &|\xi|\leq R,\\ 0, &|\xi|>R+1, \end{cases} \qquad \chi_2(\xi)= \begin{cases} 1, &|\xi|\geq R+1,\\ 0, &|\xi|<R, \end{cases} $$
where χ 1,χ 2 are smooth cut-off functions and χ 1(ξ)+χ 2(ξ)=1.

Set \(\hat{G}_{i}=\chi_{i}\hat{G}, \hat{H}_{i}=\chi_{i}\hat{H} \). For low frequencies, we have the following lemma.

Lemma 3.2

For 2≤p≤∞, we have \(\|D_{x}^{\alpha}G_{1}\|_{L_{p}}\leq Ct^{-\frac{|\alpha|}{2}-\frac{n}{2}(1-\frac{1}{p})}, \|D_{x}^{\alpha}H_{1}\|_{L_{p}}\leq Ct^{-\frac{|\alpha|}{2}-\frac{n}{2}(1-\frac{1}{p})}\).

Proof

By the Hausdorff–Young inequality, if the integers p≥2 and q satisfy \(\frac{1}{p}+\frac{1}{q}=1\), then we have
$$\begin{aligned} \bigl\| D_x^{\alpha}G_1\bigr\| _{L_p}\leq C\bigl\| \xi^{\alpha}\hat{G}_1\bigr\| _{L_q}&\leq C \biggl(\int _{|\xi|\leq R+1}|\xi|^{|\alpha|q}e^{-\frac{\eta q |\xi |^2}{1+|\xi|^2}t}\,d\xi \biggr)^{\frac{1}{q}} \\ &\leq C \biggl(\int_{|\xi|\leq R+1}|\xi|^{|\alpha|q}e^{-\frac{\eta q |\xi |^2}{1+(R+1)^2}t}\,d\xi \biggr)^{\frac{1}{q}} \\ &\leq C \bigl(t^{-\frac{n}{2}-\frac{|\alpha|q}{2}} \bigr)^{\frac{1}{q}} \\ &=Ct^{-\frac{|\alpha|}{2}-\frac{n}{2}(1-\frac{1}{p})} \end{aligned}$$
and
$$ \bigl\| D_x^{\alpha}H_1\bigr\| _{L_p}\leq C\biggl\| \xi^{\alpha}\frac{1}{1+|\xi|^2}\hat{G}_1\biggr\| _{L_q}\leq C \bigl\| \xi^{\alpha}\hat{G}_1\bigr\| _{L_q}\leq Ct^{-\frac{|\alpha|}{2}-\frac{n}{2}(1-\frac{1}{p})}. $$
 □

Next we analyze the constructions of G,H for high frequencies.

Lemma 3.3

There exist a positive constant b and functions f 1(x),f 2(x),f 3(x) such that G 2Ce bt (δ(x)+f 1(x)),H 2Ce bt (δ(x)+f 2(x)), ∇H 2Ce bt (δ(x)+f 3(x)). Here \(\|f_{1}\|_{L_{1}}<C,\|f_{2}\|_{L_{1}}<C,\|f_{3}\|_{L_{1}}<C\).

Proof

We just prove the first inequality; the proofs of the others are similar. By (3.3), if |β|≥1 then we have
$$ \bigl|D_{\xi}^{\beta}\hat{G}_2\bigr|\leq C| \xi|^{-|\beta|-1}e^{-\frac{\eta |\xi|^2}{1+|\xi|^2}t}\leq C|\xi|^{-|\beta|-1}e^{-bt}. $$
From this and Lemma 3.2 in [9], we get our result. □

Lemma 3.4

\(\|u\|_{L_{\infty}}\leq CE\).

Proof

By Young’s inequality, from (3.5) we have
$$\begin{aligned} \|u\|_{L_{\infty}} \leq&\|G_1\ast u_0 \|_{L_{\infty}}+\|G_2\ast u_0\|_{L_{\infty}} + \int _0^t\bigl\| H_1(t-s)\bigr\| _{L_{\infty}}\| \operatorname {div}f(u)\|_{L_1}\,ds \\ &{}+\int_0^t\bigl\| H_2(t-s)\ast \operatorname {div}f(u) \bigr\| _{L_{\infty}}\,ds. \end{aligned}$$
(3.6)
By Lemmas 3.2, 3.3 and the Sobolev embedding inequality, we have
$$\begin{aligned} \|G_1\ast u_0\|_{L_{\infty}}+ \|G_2\ast u_0\|_{L_{\infty}} &\leq\|G_1\ast u_0\|_{H^l}+Ce^{-bt} \bigl(\|u_0 \|_{L_{\infty}}+\|f_1\| _{L_1}\|u_0 \|_{L_{\infty}} \bigr) \\ &\leq C\|G_1\|_{L_{\infty}}\|u_0\|_{H^l}+Ce^{-bt} \|u_0\|_{H^l} \\ &\leq CE . \end{aligned}$$
(3.7)
Setting \(M(t)=\sup_{0\leq s\leq t, |\alpha|\leq 1}\|D_{x}^{\alpha}u\|_{L_{\infty}}\), we have
$$ \bigl\| D_x^{\alpha}u\bigr\| _{L_{\infty}}\leq M, \quad |\alpha|\leq1. $$
(3.8)
Next we will prove that MCE. From (3.1) we have
$$ \|u\|_{L_2}\leq CE, \qquad \int_0^t \|\nabla u\|_{L_2}^2ds\leq CE^2. $$
By Lemma 3.2, (3.1) and Hölder inequality, we have
$$\begin{aligned} \int_0^t\bigl\| H_1(t-s) \bigr\| _{L_{\infty}} \bigl\| \operatorname {div}f(u)\bigr\| _{L_1}\,ds &\leq C\int_0^t(1+t-s)^{-\frac{n}{2}} \|\nabla u\|_{L_2}\| u\|_{L_2}\,ds \\ \leq&C \biggl(\int_0^t(1+t-s)^{-n} \| u\|_{L_2}^2ds \biggr)^{\frac {1}{2}} \biggl(\int _0^t \|\nabla u\|_{L_2}^2\,ds \biggr)^{\frac{1}{2}} \\ &\leq CE \end{aligned}$$
(3.9)
By Lemma 3.3 and (3.8),
$$\begin{aligned} \int_0^t\bigl\| H_2(t-s) \ast \operatorname {div}f(u)\bigr\| _{L_{\infty}}ds& \leq\int_0^te^{-b(t-s)} \bigl(\bigl\| \operatorname {div}f(u)\bigr\| _{L_{\infty}}+\|f_2\|_{L_1}\bigl\| \operatorname {div}f(u) \bigr\| _{L_{\infty}}\bigr) \\ &\leq\int_0^te^{-b(t-s)}\|u \|_{L_{\infty}}\|\nabla u\|_{L_{\infty }}\,ds \\ &\leq C\int_0^te^{-b(t-s)}M^2\,ds \\ &\leq CM^2. \end{aligned}$$
(3.10)
By (3.6), (3.7), (3.9), and (3.10), we have
$$ \|u\|_{L_{\infty}}\leq CE+CE+CM^2. $$
Thus M(t)≤CE+CM 2(t). Because \(M(0)=\sup_{|\alpha|\leq1}\|D_{x}^{\alpha}u_{0}\|_{L_{\infty}}\leq \|u_{0}\|_{H^{l}}+\|\nabla u_{0}\|_{H^{l}}\leq CE\), this implies that M(t)≤CE; thus our result is proved. □

Lemma 3.5

\(\|u\|_{H^{l}}\leq CE\). \(\|\nabla u\|_{H^{l}}\leq CE\).

Proof

By (2.6) and the formula f(u)=(u 2,u 2,…,u 2), we can assert that
$$ \bigl\| D_x^{\alpha}f(u)\bigr\| _{L_2}\leq C\|u \|_{L_{\infty}}\bigl\| D_x^{\alpha}u\bigr\| _{L_2}. $$
(3.11)
From (1.1) we have
$$ D_x^{\alpha}\partial_tu-D_x^{\alpha} \Delta\partial_tu-\eta D_x^{\alpha}\Delta u+D_x^{\alpha}(\beta\cdot\nabla)u=-D_x^{\alpha} \operatorname {div}f(u). $$
(3.12)
By (3.12) and Hölder inequality, we have
$$\begin{aligned} &\frac{1}{2}\frac{\partial}{\partial t}\bigl\| D_x^{\alpha}u \bigr\| _{L_2}^2+\frac {1}{2}\frac{\partial}{\partial t} \bigl\| D_x^{\alpha}\nabla u\bigr\| _{L_2}^2+\eta \bigl\| D_x^{\alpha}\nabla u\bigr\| _{L_2}^2 \\ &\quad =-\int D_x^{\alpha}u D_x^{\alpha} \operatorname {div}f(u)dx \\ &\quad =\int D_x^{\alpha}f(u)\cdot D_x^{\alpha} \nabla u\,dx \\ &\quad \leq\frac{\eta}{2}\bigl\| D_x^{\alpha}\nabla u \bigr\| _{L_2}^2+C\bigl\| D_x^{\alpha}f(u) \bigr\| _{L_2}^2. \end{aligned}$$
(3.13)
From (3.11), (3.13) and Lemma 3.4, we have
$$\begin{aligned} \frac{1}{2}\frac{\partial}{\partial t}\bigl\| D_x^{\alpha}u \bigr\| _{L_2}^2+\frac {1}{2}\frac{\partial}{\partial t} \bigl\| D_x^{\alpha}\nabla u\bigr\| _{L_2}^2+ \frac{\eta}{2}\bigl\| D_x^{\alpha} \nabla u\|_{L_2}^2 \leq& C\bigl\| D_x^{\alpha}f(u)\bigr\| _{L_2}^2 \\ \leq& C\|u\|_{L_{\infty}}^2\bigl\| D_x^{\alpha}u \bigr\| _{L_2}^2 \\ \leq& CE^2\bigl\| D_x^{\alpha}u\bigr\| _{L_2}^2. \end{aligned}$$
Then
$$\begin{aligned} &\bigl\| D_x^{\alpha}u\bigr\| _{L_2}^2+ \bigl\| D_x^{\alpha}\nabla u\bigr\| _{L_2}^2+ \frac{\eta }{2}\int_0^t\bigl\| D_x^{\alpha} \nabla u\bigr\| _{L_2}^2 \\ &\quad \leq CE^2\int_0^t \bigl\| D_x^{\alpha}u\bigr\| _{L_2}^2ds+ \bigl\| D_x^{\alpha}u_0\bigr\| _{L_2}^2+ \bigl\| D_x^{\alpha}\nabla u_0\bigr\| _{L_2}^2. \end{aligned}$$
By (3.1), if |α|=1 then
$$ \bigl\| D_x^{\alpha}u\bigr\| _{L_2}\leq CE,\qquad \int _0^t\bigl\| D_x^{\alpha}\nabla u \bigr\| _{L_2}^2ds\leq CE^2. $$
Using induction, we have
$$ \bigl\| D_x^{\alpha}u\bigr\| _{L_2}\leq CE,\qquad \bigl\| D_x^{\alpha}\nabla u\bigr\| _{L_2}\leq CE,\quad|\alpha| \leq l. $$
(3.14)
Thus our lemma is proved. □

According to Lemma 3.5 and using the local solution, by the usual method, we can derive a global solution for (1.1) such that uL (0,∞,H l (R n )).

4 Decay estimation

Setting \(M(t)=\sup_{|\alpha|\leq l, 0\leq s\leq t}\|D_{x}^{\alpha}u\|_{L_{2}}(1+s)^{\frac{n}{4}+\frac{|\alpha|}{2}}\), we have
$$ \bigl\| D_x^{\alpha}u\bigr\| _{L_2}\leq M(t) (1+s)^{-\frac{n}{4}-\frac{|\alpha|}{2}},\quad|\alpha|\leq l. $$
(4.1)
By (4.1),
$$ \bigl\| D_x^{\alpha}f(u) (s)\bigr\| _{L_1}\leq\sum _{|\alpha_1|+|\alpha_2|=|\alpha|}\bigl\| D_x^{\alpha_1}u \bigr\| _{L_2}\bigl\| D_x^{\alpha _1}u\bigr\| _{L_2} \leq CM^2(t) (1+s)^{-\frac{n}{2}-\frac{|\alpha|}{2}}. $$
(4.2)
From (3.11), Lemma 3.4 and (4.1), it follows that
$$ \bigl\| D_x^{\alpha}f(u)\bigr\| _{L_2}\leq C\|u \|_{L_\infty}\bigl\| D_x^{\alpha}u\bigr\| _{L_2}\leq CEM(t) (1+s)^{-\frac{n}{4}-\frac{|\alpha|}{2}},\quad |\alpha|\leq l. $$
(4.3)
By (3.4) and Young’s inequality,
$$\begin{aligned} \bigl\| D_x^{\alpha}u\bigr\| _{L_2} \leq& \bigl\| D_x^{\alpha}G_1\bigr\| _{L_2} \|u_0\|_{L_1}+\bigl\| G_2\ast D_x^{\alpha}u_0 \bigr\| _{L_2} \\ &{}+ \int_0^t\bigl\| D_x^{\alpha}H_1(t-s) \ast \operatorname {div}f(u)\bigr\| _{L_2}ds+\int_0^t\bigl\| \nabla H_2(t-s)\ast D_x^{\alpha}f(u) \bigr\| _{L_2}\,ds. \end{aligned}$$
(4.4)
From Lemmas 3.2 and 3.3, we have
$$ \bigl\| D_x^{\alpha}G_1\bigr\| _{L_2} \|u_0\|_{L_1}+\bigl\| G_2\ast D_x^{\alpha}u_0 \bigr\| _{L_2}\leq CE(1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}},\quad|\alpha|\leq l. $$
(4.5)
When |α|≤l, from Lemma 3.2, (4.2) and (3.1), we get
$$\begin{aligned} &\int_0^t\bigl\| D_x^{\alpha}H_1(t-s) \ast \operatorname {div}f(u)\bigr\| _{L_2}\,ds \\ &\quad \leq\int_0^{\frac{t}{2}} \bigl\| D_x^{\alpha}\nabla H_1(t-s)\bigr\| _{L_2} \bigl\| f(u)\bigr\| _{L_1}\,ds \\ &\qquad{}+\int_{\frac{t}{2}}^t \bigl\| \nabla H_1(t-s) \bigr\| _{L_2}\bigl\| D_x^{\alpha} f(u)\bigr\| _{L_1}\,ds \\ &\quad \leq\int_0^{\frac{t}{2}}(1+t-s)^{-\frac{n}{4}-\frac{|\alpha|}{2}-\frac {1}{2}}M^2(1+s)^{-\frac{n}{2}}\,ds \\ &\qquad{}+ \int_{\frac{t}{2}}^t(1+t-s)^{-\frac{n}{4}-\frac{1}{2}}M^2(t) (1+s)^{-\frac{n}{2}-\frac{|\alpha|}{2}}\,ds \\ &\quad \leq CM^2(1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}+C(1+t)^{-\frac{n}{4}-\frac {|\alpha|}{2}}M^2(t). \end{aligned}$$
(4.6)
From Lemma 3.3 and (4.3), with |α|≤l, we have
$$\begin{aligned} &\int_0^t\bigl\| \nabla H_2(t-s)\ast D_x^{\alpha} f(u)\bigr\| _{L_2}\,ds \end{aligned}$$
(4.7)
$$\begin{aligned} &\quad \leq \int_0^te^{-b(t-s)}\bigl( \bigl\| D_x^{\alpha} f(u)\bigr\| _{L_2} +\|f_3 \|_{L_1}\bigl\| D_x^{\alpha} f(u)\bigr\| _{L_2}\bigr)\,ds \\ &\quad \leq \int_0^te^{-b(t-s)}CEM(t) (1+s)^{-\frac{n}{4}-\frac{|\alpha |}{2}}\,ds \\ &\quad \leq CEM(t) (1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}. \end{aligned}$$
(4.8)
From (4.4)–(4.7) we obtain
$$ \bigl\| D_x^{\alpha}u \bigr\| _{L_2}\leq CE(1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}+CM^2(1+t)^{-\frac{n}{4}-\frac {|\alpha|}{2}}+ CEM(t) (1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}, $$
thus M(t)≤CE+CM 2(t)+CEM(t). In addition, since M(0)≤CE, we have M(t)≤CE. Therefore,
$$ \bigl\| D_x^{\alpha}u\bigr\| _{L_2}\leq CE(1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}. $$
(4.9)
Thus we get Theorem 1.1, our main result in this paper.

Notes

Acknowledgements

The first author was supported by NNSF 11101121.

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Authors and Affiliations

  1. 1.College of ScienceHohai UniversityNanjingChina
  2. 2.China Electronics Technology Group Corporation No. 28 Research InstituteNanjingChina

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